Check if given two Arrays are equal (using Map)
Last Updated :
08 Sep, 2022
Given two arrays, A[] and B[], the task is to check if they are equal or not. Arrays are considered equal if any permutation of array B equates to array A.
Examples:
Input: A[] = [2, 4, 5, 7, 5, 6] and B[] = [4, 2, 5, 5, 6, 7]
Output: Yes
Explanation: All the elements in array A are present in array B and same number of times.
Input: A[] = [2, 5, 8, 9, 78] and B[] = [5, 2, 7, 78, 8]
Output: No
Explanation: In array A there is a 9 and in array B there is a 7
Approach: The problem can be solved using hashmap based on the below idea:
Two arrays will be equal only if the frequency of the respective elements in both arrays are equal.
Follow the steps to solve the problem:
- If the sizes of both arrays are not equal, return NO.
- Maintain a hashmap and count the frequency of both the array elements
- If for any element the frequency is not the same, then return NO
- Otherwise, return YES
Below is the implementation of the above approach.
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
bool isEqual(vector<int> a, vector<int> b, int n, int m)
{
// If size is not same return false
if (n != m) {
return 0;
}
// Create 2 unordered maps to store
// the frequency
unordered_map<int, int> mp1, mp2;
for (int i : a) {
mp1[i]++;
}
for (int i : b) {
mp2[i]++;
}
// Compare the frequency
for (auto i = mp1.begin(); i != mp1.end(); i++) {
// If frequency not same return false
if (mp2[i->first] != i->second) {
return 0;
}
}
return 1;
}
// Drivers code
int main()
{
vector<int> a = { 2, 4, 5, 7, 5, 6 },
b = { 4, 2, 5, 5, 6, 7 };
int n = a.size(), m = b.size();
bool flag = isEqual(a, b, n, m);
// Return 1 if equal
if (flag) {
cout << "YES\n";
}
else {
cout << "NO\n";
}
return 0;
}
// Java code to implement the approach
import java.io.*;
import java.util.*;
class GFG {
public static boolean isEqual(int a[], int b[], int n,
int m)
{
// If size is not same return false
if (n != m) {
return false;
}
// Create 2 unordered maps to store
// the frequency
HashMap<Integer, Integer> mp1
= new HashMap<Integer, Integer>();
HashMap<Integer, Integer> mp2
= new HashMap<Integer, Integer>();
for (int i : a) {
if (mp1.get(i) != null)
mp1.put(i, mp1.get(i) + 1);
else
mp1.put(i, 1);
}
for (int i : b) {
if (mp2.get(i) != null)
mp2.put(i, mp2.get(i) + 1);
else
mp2.put(i, 1);
}
// Compare the frequency
for (Map.Entry<Integer, Integer> i :
mp1.entrySet()) {
Integer key = i.getKey();
// If frequency not same return false
if (mp2.get(key) != i.getValue()) {
return false;
}
}
return true;
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 2, 4, 5, 7, 5, 6 };
int b[] = { 4, 2, 5, 5, 6, 7 };
int n = a.length, m = b.length;
boolean flag = isEqual(a, b, n, m);
// Return 1 if equal
if (flag == true) {
System.out.print("YES\n");
}
else {
System.out.print("NO\n");
}
}
}
// This code is contributed by Rohit Pradhan
# Python3 code to implement the approach
def isEqual(a, b, n, m) :
# If size is not same return false
if (n != m) :
return 0
# Create 2 unordered maps to store
# the frequency
mp1 = dict.fromkeys(a, 0);
mp2 = dict.fromkeys(b, 0);
for i in a :
if i in mp1 :
mp1[i] += 1;
else:
mp1[i] = 0
for i in b :
if i in mp2 :
mp2[i] += 1;
else :
mp2[i] = 0
# Compare the frequency
for i in mp1 :
# If frequency not same return false
if (mp2[i] != mp1[i]) :
return 0;
return 1;
# Drivers code
if __name__ == "__main__" :
a = [ 2, 4, 5, 7, 5, 6 ];
b = [ 4, 2, 5, 5, 6, 7 ];
n = len(a);
m = len(b);
flag = isEqual(a, b, n, m);
# Return 1 if equal
if (flag) :
print("YES");
else :
print("NO");
# This code is contributed by AnkThon
// C# code to implement the approach
using System;
using System.Collections.Generic;
public class GFG {
public static bool isEqual(int []a, int []b, int n,
int m)
{
// If size is not same return false
if (n != m) {
return false;
}
// Create 2 unordered maps to store
// the frequency
Dictionary<int, int> mp1
= new Dictionary<int, int>();
Dictionary<int, int> mp2
= new Dictionary<int, int>();
foreach (int i in a) {
if (mp1.ContainsKey(i))
mp1[i] = mp1[i] + 1;
else
mp1.Add(i, 1);
}
foreach (int i in b) {
if (mp2.ContainsKey(i))
mp2[i] = mp2[i] + 1;
else
mp2.Add(i, 1);
}
// Compare the frequency
foreach (KeyValuePair<int, int> i in
mp1) {
int key = i.Key;
// If frequency not same return false
if (mp2[key] != i.Value) {
return false;
}
}
return true;
}
// Driver Code
public static void Main(String[] args)
{
int []a = { 2, 4, 5, 7, 5, 6 };
int []b = { 4, 2, 5, 5, 6, 7 };
int n = a.Length, m = b.Length;
bool flag = isEqual(a, b, n, m);
// Return 1 if equal
if (flag == true) {
Console.Write("YES\n");
}
else {
Console.Write("NO\n");
}
}
}
// This code is contributed by shikhasingrajput
<script>
// Javascript code to implement the approach
function isEqual(a, b, n, m)
{
// If size is not same return false
if (n != m) {
return false;
}
// Create 2 unordered maps to store
// the frequency
let mp1 = new Map();
let mp2 = new Map();
for (let i of a) {
if (mp1.get(i) != null)
mp1.set(i, mp1.get(i) + 1);
else
mp1.set(i, 1);
}
for (let i of b) {
if (mp2.get(i) != null)
mp2.set(i, mp2.get(i) + 1);
else
mp2.set(i, 1);
}
// Compare the frequency
for (let i of mp1.keys()) {
// If frequency not same return false
if (mp2.get(i) != mp1.get(i)) {
return false;
}
}
return true;
}
// Driver Code
let a = [2, 4, 5, 7, 5, 6];
let b = [4, 2, 5, 5, 6, 7];
let n = a.length, m = b.length;
let flag = isEqual(a, b, n, m);
// Return 1 if equal
if (flag == true) {
document.write("YES<br>");
}
else {
document.write("NO<br>");
}
// This code is contributed by Saurabh Jaiswal
</script>
Time complexity: O(N) in average case and O(N2) in worst case.
Auxiliary Space: O(N)
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